INFLUENCE  DTAGEAMS 


FOR    THE 


DETERMINATION    OF    MAXIMUM 
MOMENTS  IN  TRUSSES  AND  BEAMS 


BY 

MALVEED  A.    HOWE,   C.E. 

PROFESSOR  OF  CIVIL  ENGINEERING,   ROSE   POLYTECHNIC  INSTITUTE 
MEMBER   AMERICAN   SOCIETY    OF   CIVIL  ENGINEERS 


FIRST  EDITION 
FIRST    THOUSAND 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC. 

LONDON:     CHAPMAN    &   HALL,   LIMITED 

1914 


Copyright,  19U 

BY 

MALVERD  A.  HOWE 


THE  SCIENTIFIC  PRESS 

ROBERT   DRUM MONO    AND  COMPANY 

BROOKLYN,  N.   Y. 


PREFACE 


THE  use  of  influence  lines  for  the  determination  of 
maximum  moments  and  criterions  for  wheel  loads  is  pre- 
sented in  many  modern  text  books. 

The  object  of  these  few  pages  is  to  bring  attention  to 
the  fact  that  for  loads  on  all  ordinary  trusses,  the  influence 
diagrams  for  bending  moments  are  drawn  by  following 
a  single  simple  rule,  and  that  the  diagrams  so  con- 
structed require  no  computations  for  their  direct  applica- 
tion. In  addition  to  this  the  influence  diagrams  for  loads 
on  continuous  trusses,  cantilever  trusses  and  arches  are 
shown  to  be  based  upon  the  one  general  diagram  for  simple 
trusses. 

While  the  diagrams,  as  a  rule,  are  constructed  for 
moments  yet  they  can  be  as  easily  drawn  for  stresses  or 
even  areas  of  truss  members. 

The  use  of  the  influence  diagrams  in  the  determination 
of  criterions  for  the  positions  of  wheel  loads  which  produce 
maximums  is  explained  and  shown  to  be  very  simple. 

M.  A.  H. 

DECEMBER,  1913. 

iii 


TABLE   OF   CONTENTS 


CHAPTER  I 

SIMPLE  TRUSSES 

PAGE 

Rule  for  Drawing  Influence  Diagrams 2 

Diagonal  of  Pratt  Truss 3 

Load  Position  for  Maximum  Moment 5 

Neutral  Point 7 

Vertical  of  Pratt  Truss 8 

Web  Member  of  Warren  Truss 9 

Web  Members  in  Simple  Trusses  having  Parallel  Chords 9 

Bottom  Chord  of  Curved  Chord  Simple  Truss 12 

Top  Chord  of  Curved  Chord  Simple  Truss 13 

Bottom  Chord  of  Curved  Chord  Warren  Truss 14 

Trusses  with  Sub-divided  Panels 15 

Lower  Segment  of  Diagonal  in  Sub-strut  Truss 15 

Upper  Segment  of  Diagonal  in  Sub-strut  Truss 17 

Vertical  of  Sub-strut  Truss 18 

Upper  Segment  of  Diagonal  in  Sub-hanger  Truss 19 

Lower  Segment  of  Diagonal  in  Sub-hanger  Truss 20 

Vertical  of  Sub-hanger  Truss 21 

Bottom  Chord  of  Sub-strut  Truss 21 

Top  Chord  of  Sub-strut  Truss 23 

Top  Chord  of  Sub-hanger  Truss 24 

Bottom  Chord  of  Sub-hanger  Truss 25 

The  Length  of  the  Cut  Stringer 26 

Diagonal   of  Simple  Truss  having  Center  of  Moments  between  the 

Supports 27 

Vert  ical  of  Simple  Truss  having  Center  of  Moments  between  the  Supports .  28 

v 


vi  TABLE  OF  CONTENTS 

CHAPTER  II 
DOUBLE  INTERSECTION  TRUSSES 

PAGE 

Double  Intersection  Trusses 30 

Top  Chord  of  Whipple  Truss 30 

Vertical  of  Whipple  Truss 31 

Diagonal  of  Whipple  Truss 32 

Top  Chord  of  Sub-divided  Double  Triangular  Truss 32 

Bottom  Chord  of  Sub-divided  Double  Triangular  Truss 33 

Lower  Segment  of  Diagonal  in  Sub-divided  Double  Triangular  Truss ...  35 

Chords  of  the  K  Truss 35 

Diagonal  Web  Members  of  the  K  Truss 36 

Upper  Segment  of  Vertical  in  the  K  Truss 38 

Lower  Segment  of  Vertical  in  the  K  Truss 39 

CHAPTER  III 

CONTINUOUS  TRUSSES 

The  Moment  Influence  Diagrams 41 

General  Moment  Formula 41 

Cantilever  Truss 43 

Top  Chord  of  Anchor  Span 43 

Top  Chord  of  Cantilever  Span 44 

Diagonal  of  Anchor  Span 46 

Diagonal  of  Cantilever  Span 47 

Partially  Continuous  Truss 48 

Top  Chord  of  a  Partially  Continuous  Truss 49 

Diagonal  of  a  Partially  Continuous  Truss 52 

Continuous  Truss  of  Two  Equal  Spans 53 


CHAPTER  IV 
ARCHES 

Arches 55 

Diagonals  of  a  Three-hinged  Arch  with  Open  Web 55 

Chord  of  a  Two-hinged  Arch  with  Open  Web 57 

Diagonal  of  a  Two-hinged  Arch  with  Open  Web 58 


TABLE  OF  CONTENTS  vii 


Moment  Influence  Diagram  for  Two-hinged  Arch  with  Solid  Web 59 

Vertical  Shear  for  Two-hinged  Arch  with  Solid  Web 59 


CHAPTER  V 
BEAMS  OF  CONSTANT  CROSS-SECTION 

The  Influence  Diagram 60 

Restrained  Beams 60 

Simple  Beam  on  Two  Supports 61 

Beam  Fixed  at  One  End  and  Supported  at  the  Other  End 62 

Beam  Fixed  at  Both  Ends 63 


DEFINITION 

An  influence  diagram  is  one  which  shows  the  effect 
of  a  unit  load  moving  across  a  structure  upon  any  function 
of  the  structure  for  any  position  of  the  load. 

viii 


INFLUENCE  DIAGRAMS  FOR  THE  DETERMINATION 
OF  MAXIMUM  MOMENTS 


CHAPTER  I 

SIMPLE    TRUSSES* 

An  Influence  Line  for  the  bending  moments  at  the 
center  of  moments  for  any  member  of  a  truss  on  two  sup- 
ports for  vertical  loads  can  be  constructed  by  a  method 
which  is  perfectly  general  in  its  application.  The  truss 
may  be  of  any  shape,  and  the  loads  may  be  applied  to 
either  the  upper-  or  lower-chord  joints. 

A  section  is  passed  through  the  truss,  in  the  usual 
manner,  cutting  the  member  whose  influence  line  is  to  be 
drawn.  This  section  must  evidently  also  cut  a  stringer 
in  the  panel  of  the  truss  containing  the  cut  loaded  chord. 
Let: 

d=the  horizontal  projection  of  the  length  of  the  stringer 
which  is  cut  by  the  section.  This  stringer  will  be 
called  the  cut  stringer.  ^}q  c 

b  =  the  horizontal  distance  from  the  left  support  of  the 
truss  to  the  left  end  of  the  cut  stringer. 

*  See  General  Method  for  Drawing  Influence  Lines  for  Stress  in  Simple 
Trusses,"  by  Malverd  A.  Howe.  Engineering  News,  June  12.  1913. 


2  INFLUENCE  DIAGEAMS 

s  =  the  horizontal  distance  from  the  left  support  of  the 
truss  to  the  center  of  moments,  s  will  be  considered 
positive  when  measured  to  the  right. 

h  =  the  vertical  distance,  at  the  left  end  of  the  cut  stringer, 
between  the  two  chord  members  which  are  cut  by 
the  section. 
hf  =  the  vertical  distance  between  the  same  chords  at  the 

right  end  of  the  cut  stringer. 

The   General   Rule  for  drawing  the  influence  line  is  as 
follows:  % 

(a)  Through  any  point  A  (refer  to  Fig.  1;  the  other 
figures  are  lettered  correspondingly)  in  a  vertical  line  pass- 
ing through  the  center  of  moments^  draw  a  horizontal 
line  cutting  the  vertical  line  through  -the  left  support  of 
russ  at  C  and  the  vertical  line  through  the  right  support 


(6)  From  C,  at  any  convenient  scale,  lay  off  vertically 
downward  the  distance  CD=AC  =s,  and  connect  B  and  D 
by  a  straight  line. 

(c)  Through  A  and  D  draw  a  straight  line  and  prolong 
it  until  it  cuts  the  vertical  line  drawn  through  the  left  end 
of  the  cut  stringer  at  E. 

(d)  Draw  a  vertical  line  through  the  right  end  of  the 
cut  stringer,  intersecting  the  line  ACB  at  F,  and  connect 
E  and  F  by  a  right  line. 

(e)  The  polygon  DEFBD  contains  the  influence  line 
sought;    the  ordinates  between  the  line  DEFB  and  the 
line  DB  are  the  respective  moments  for  unit  loads  on  the 
truss  vertically  above  them,  i.e.,  the  lines  ef=Zi,  rc=z2, 
gk=zz,  are  proportional  to  the  stresses  produced  by  loads 
of   one    pound   at   Wi,    Wz   and    Ws,   respectively.     The 
ordinates  are  measured  at  the  scale  used  in  laying  off  the 
distance  CD=s. 

The  application  of  the  above  rule  to  several  forms  of 
trusses  will  now  be  considered. 


SIMPLE  TRUSSES  3 

Diagonal  of  Pratt  Truss  with  Inclined  Top  Chord.— 
Fig.  1  shows  this  truss  and  the  influence  diagram  for  the 
member  UL'.  The  shaded  area  is  the  influence  diagram. 

To  prove  the  construction  correct,  let  the  angles  made 
with  AB  by  the  lines  AE,  EF  and  DB  be  respectively 
0i,  62  and  03,  and  let  the  distance  from  the  left  support 
of  the  truss  to  the  unit  load  be,  in  general,  represented  by 


'*   Criterion,  =(Wi  +  HY) 


FIG.  1. 

a,     While  the  load  W\  is  to  the  left  of  the  cut  stringer,  or 
to  the  left  of  the  vertical  through  E,  the  moment  is: 


4,  d) 


or 


tan0i,      .     (2) 
=  TTi  ( -  pe +fp)  =  Wl  ( +ef)  .  (3) 


4  INFLUENCE  DIAGRAMS 

Since  the  resultant  moment  is  positive,  the  moment  UL'(y) 
is  negative  and  therefore  UL'  is  in  compression. 

If  the  load,  W2  (or  W2f)  is  on  the  cut  stringer,  or  between 
the  verticals  through  E  and  F,  the  moment  is,  for  W2, 


.     .     (4) 
-a2f-^~,.     .     (5) 

d 

< 

or 

M8=  -W2(l-a2)  tan03  +  TF2(&+d-a2)tan02;  (6) 
and  hence 

Ms  =  W2(-qr+qc)=W2(+rc)      .....     (7) 

Since  the  resultant  moment  is  positive  the  stress  in  UL' 
is  compressive. 

For  the  load  TF2' 

Ms=W2'(-rc),     ......     (8) 

and  UL'  is  in  tension. 

When  the  load  is  between  the  cut  stringer  and  the  right 
support  of  the  truss,  the  moment  is: 

M,=  -W^^s=  -Wz(l-a3)S  =  -TF3(Z-a3)tan03, 


or 

Ms  =  W3(-gk),     .    .....     (9) 

and  the  member  UL'  is  in  tension. 


SIMPLE  TRUSSES  5 

If  the  diagonal  inclines  in  the  opposite  direction,  as  in 
a  Howe  truss,  the  construction  of  the  influence  diagram 
remains  unchanged  but  the  character  of  the  stress  is  reversed. 

Load-position  for  Maximum  Moment.  —  For  uniform 
loads,  the  influence  diagram  DEGFB  at  once  indicates  the 
portions  of  the  span  which  are  loaded  to  produce  like 
moments  and  hence  maximum  stresses. 

The  criterion  for  giving  the  position  of  wheel  loads 
producing  the  maximum  moment  is  readily  found  from  the 
influence  diagram.  For  cases  which  usually  occur  in 
practice,  the  portion  of  the  span  on  the  left  of  the  cut 
stringer  may  be  considered  as  unloaded.  For  convenience  W* 
is  assumed  to  represent  all  of  the  loads  between  the  verticals 
through  E  and  G  concentrated  at  their  center  of  gravity, 
WY  all  of  the  loads  between  the  verticals  through  G  and  F, 
and  Ws  all  of  the  loads  between  the  verticals  through  F 
and  B.  Let  22,  22'  and  23,  be  the  ordinates  of  the  influence 
diagram  directly  below  the  wheel  loads  W2,  W%  and  TF3. 
The  moment  is 

Ms  =  W2(z2)-W2'(z2')-W3(z3).   .     .     .     (10) 

If  the  loads  move  toward  the  left  a  distance  dx,  and  no 
additional  load  comes  on  the  span  from  the  right  and  no 
load  moves  off  the  left  end  of  the  cut  stringer,  the  moment 
becomes : 

Ml  =  ^2(22  -  dx  tan  03  +  dx  tan  02) 

-TF2'(z2'  +  dx  tan  03  -  dx  tan  02)  -W3(z3  +  dx  tan  03).     (11) 

The  difference  between  these  moments  is, 

Ms'  -Ma  =  dMs  =  W2(  -  tan  03  +tan  02)  dx 

-TF2'(tan  03-tan  e2)dx-W3  (tan  03)5o;.     (12) 


6  INFLUENCE  DIAGRAMS 


Dividing  through  by  5x  and  placing  — -  =  0, 


(13) 

(14) 


where  W  =  W2+  W2'  +  W3  =  the  total  load  on  the  span. 
Therefore,  from  equation  (14),  the  desired  criterion  is 


W 

.     •     •     (15) 


where 


s+b  d 

--    or    d 


The  value  of  d'  is  found  graphically  from  the  influence 
diagram  by  drawing  a  line  through  D  parallel  to  CB  until 
it  cuts  the  vertical  line  through  F,  then  d!  is  the  distance 
indicated  in  the  figure  (Fig.  1  and  those  which  follow). 
This  value  may  also  be  found  graphically  without  drawing 
the  influence  diagram  by  simply  drawing  in  the  truss  dia- 
gram the  line  Mm  parallel  to  the  bottom  cut  chord  LU  and 
prolonging  it  until  it  cuts  a  diagonal  line  drawn  from  the 
intersection  of  the  cut  top  chord  and  a  vertical  through  the 
left  end  of  the  cut  stringer  to  the  intersection  of  the  cut 
bottom  chord  and  the  vertical  through  the  right  end  of 
the  cut  stringer.  This  point  is  indicated  by  the  letter  m  in 
the  figure.  The  horizontal  distance  of  this  point  m  from 


SIMPLE  TRUSSES  7 

the  vertical  through  the  right  end  of  the  cut  stringer  is  the 
value  of  d'.     This  is  easily  shown  as  follows:    Fig.  1. 

mn  :  h  ::  d'  :  d,     or  df  =  j-mn  =  -rMN]  .     .     (17) 
but 

MNisiih  :s+b,   or  MN  =-^h;    .     .     .     (18) 
therefore 

...'..    (19) 


which  is  the  value  given  in  (16). 

Neutral  Point.  —  The  position  of  a  load  which  produces 
no  stress  in  the  web  member  UL'  is  indicated  by  the  point 
G  in  the  influence  diagram,  Fig.  1,  since  a  load  in  the  truss 
immediately  above  G  produces  no  stress  in  the  member 
as  shown  by  a  zero  ordinate  in  the  influence  diagram. 
The  point  w'shown  in  the  truss  diagram  is  directly  above 
G.  This  point  is  located  by  the  intersection  of  the  line 
M  Nf  and  the  diagonal  drawn  as  explained  in  determining 
d'.  Let  x'  be  the  horizontal  distance  from  mf  to  the  vertical 
through  the  right  end  of  the  cut  stringer,  and  H-}-xf  the 
horizontal  distance  of  m'  from  the  vertical  through  the 
right  support  of  the  truss.  Then  in  Fig.  1 

m'n'  :h::x'  :  d,     or    m'ri'  =  ^x'  =  MN(H+x')\;    (20) 

Ci  i 

but 

MN  :s::h  :  s+6,     or    MN  =         h;      ....     (21) 


8 
hence 


and 


INFLUENCE  DIAGRAMS 


mn  = 


KA-V,    ....   (22) 


.     .     (23) 


this  becomes 


(H+xr)  tan  03  =z'  tan  82. 


(24) 


The  ordinate  above  the  point  G  in  the  influence  diagram 
satisfies  this  equality,  and  therefore  G  and  ra'  are  in  the 
same  vertical  line. 


^.     Tan  63  =4- 
a  < 


FIG.  2. 


Vertical  of  Pratt  Truss  with  Inclined  Top  Chord.— The  truss 
diagram  and  the  influence  diagram  are  shown  in  Fig.  2. 


SIMPLE  TRUSSES  9 

All  of  the  demonstrations  given  for  Fig.  1  apply  to  Fig.  2. 
For  all  loads  upon  the  left  of  the  vertical  through  G  the 
vertical  is  in  tension  and  for  those  upon  the  right  of  G 
it  is  in  compression. 

Web  Member  of  Warren  Truss  with  Inclined  Top  Chord. 
— The  web  member  U'L,  Fig.  3  has  been  selected.  A 
comparison  with  Fig.  1  shows  that  the  influence  diagram 


/  Criterion.      -      =  (PF2  +  HY)~  = 
sd 


FIG.  3. 


is  constructed  in  exactly  the  manner  followed  in  Fig.  1, 
and  that  all  of  the  demonstrations  of  Fig.  1  apply  equally 
well  to  Fig.  3.  The  locations  of  points  m  and  m'  are 
clearly  shown  in  the  figure  where  the  verticals  h  and  hr 
and  the  diagonal  UL'  are  not  members  of  the  truss. 

Web  Members  in  Simple  Trusses  Having  Parallel  Chords. 

—The  case  shown  in  Fig.  4,  presents  an  apparent  exception 

to  the  general  rule  for  drawing  the  influence  line  DEFB. 

The  center  of  moments  for  any  web  member  of  a  truss 

having  parallel  chords  is  at  infinity.     Taking  N,  the  left 


10 


INFLUENCE  DIAGRAMS 


support  of  the  truss,  as  a  reference  point  s  =  oc.     For  a 
load  on  the  left  of  the  vertical  through  Ey  the  moment  is 


M,=  _}Ti 


+T7i(oo+ai).  (25) 


Dividing  both  members  of  equation  (25)  by  +00. 


(26) 


But,—  Wi — j-^  +  Wi  is  the  expression  f or « vertical  shear 

on  the  right  of  TFi,  and,  consequently,  practically  at  the 
load  Wi. 


If  CD  is  laid  off  equal  to  s^-oo  =1,  the  ordinates  of  the 
influence  diagram  represent  vertical  shears  instead  of 
moments.  DE  drawn  through  A,  which  is  an  infinite 


SIMPLE  TRUSSES  11 

distance  from  N,  will  be  sensibly  parallel  to  ACB.  It 
appears  that  the  vertical  shear  influence  diagram  is  con- 
structed in  precisely  the  manner  outlined  for  the  moment 
influence  diagram  when  CD  is  made  equal  to  unity. 

For  a  load  on  the  left  of  the  vertical  through  E,  the 
vertical  shear  between  E  and  F  is 


Z-ai)-TTi(l);.     (27) 


or 

-S=+Wi(l-a1)tui05-Wi(i)=Wi(+pe-pf)-, 


therefore  +S  =  Wi(+ef).     .     .     .     (28) 

In  a  similar  manner  it  is  shown  that  the  ordinates 
directly  below  W^  Wz  and  T7s,  represent  the  vertical 
shears  produced  by  unit  loads  at  these  points. 

A  negative  ordinate  in  the  influence  diagram  indicates 
that  the  resultant  shear  acts  upward;  therefore  the 
vertical  component  of  the  stress  in  the  web  member,  for 
which  the  influence  diagram  is  drawn,  acts  downward.  In 
Fig.  4,  the  member  U'L'  is  in  compression  for  loads  on 
the  left  of  G.  The  vertical  UL  has  the  same  influence 
diagram  but  it  is  in  tension  for  the  same  loads. 

The  criterion  for  maximum  shear  is  determined  in  the 
manner  given  above  for  maximum  moment.  Without 
writing  all  of  the  equations  and  referring  to  equation  (13} 
and  Fig.  4, 


~Cf 

-tan  05+tan  04)-TCY(-tan  04+tan  05) 


dx 

-W3  tan  02=0 (29) 


12 
or 


which  becomes 


INFLUENCE  DIAGRAMS 


tan  05  = 


I 


d' 


2')  tan  04,  .     (30) 


(6L) 


The  points  m  and  m'  and  the  value  of  d'  are  found  in  the 
manner  explained  for  Figs,  1,  2  and  3. 

Bottom  Chord  of  Curved-chord  Simple  Truss. — The  influence 
diagram  for  this  case  is  constructed  according  to  the  general 
directions  as  shown  in  Fig.  5.  The  points  E  and  A  coincide. 


~  .,     .  W     Wi     Wi 

Criterion.     —  = —  =-:— . 

ISO 

Tan  0i=l.      Tan  8t=*-^.      Tan0j=4- 
a  I 

FIG.  5. 


The  positive  sign  indicates  that  the  resultant  moment  is 
positive  and  that  the  moment  of  the  stress  in  LU  is  negative. 
This  shows  that  LU  is  in  tension. 

The  criterion  for  the  maximum  moment  when  wheel 
loads  are  on  the  bridge  is  found  as  follows:  (see  equation 
13). 


8MS 
6x 


Wi(  -tan  0i  +tan  03)  +W2  tan  03+TF3  tan  03  =0     (32) 


=  (TFi+T72+TF3)  tan  03-TFi  tan  0i=0; 


(33) 


SIMPLE  TRUSSES 


this  becomes 


13 


(34) 


An  equivalent  criterion  is 


l-s 


_  _ ;  __  •  •    •"    '    •  •    o  _•_• i  /n  K\ 

:  s          AB      -AT     •     •     W 


W\ 
AC' 


This  criterion  holds  good  for  any  influence  diagram 
which  is  triangular  in  shape. 

Top  Chord  of  Curved-chord  Simple  Truss. — The  construc- 
tion of  the  influence  diagram  follows  the  same  general  rule 


uu 


Tan  01=1.     Tan  02=^.     Tan  fc  =^-. 


FIG.  6. 

as  in  the  previous  case.    The  points  A  and  F,  Fig.  6  coincide, 
while  E  is  on  the  right  line  connecting  D  and  A  since  the 
line  EF  coincides  with  this  line. 
For  wheel  loads  the  criterion  is 


•W1+W2     W3     W 

—    =  ^=~'      •    •    '    •     (36) 


where  W  =  the  total  load  on  the  span. 


14 


INFLUENCE  DIAGRAMS 


Bottom  Chord  of  Curved-chord  Warren  Truss.— If  both  sets 
of  web  members  are  inclined  as  shown  in  Fig.  7,  the 
influence  diagram  remains  unchanged  in  its  construction, 
but  the  line  EF  does  not  coincide  with  the  line  DA  as  in 
the  previous  case. 


Criterion.     ~  =— '  +  W*— - . 
I        a  sd 


Tan  ft=l.      Tan 

FIG.  7. 


-     Tan* -. 


The  criterion  for  the  maximum  moment  produced  by 
wheel  loads  is  found  as  follows : 


8M, 

6x 


-tan  0i+tan  03)+TF2(-tan  02+tan 


tan  03=0.     (37) 


(Wi+W2+W3)  tan  03-TFi  tan 


-W2  tan  02=0,     (38) 

Substituting    the    values  of   the    tangents,    equation    (38) 
becomes 


=0.     .     .     (39) 


SIMPLE  TRUSSES  15 

Hence 


If  s  —  b  =  ^d,  then 

-          %-*+&*  .......     (41) 

Trusses  with  Sub-divided  Panels.  —  The  usual  type  of  truss 
with  sub-divided  panels  has  one  set  of  web  members  vertical. 
The  trusses  here  considered  will  have  one  set  of  web  mem- 
bers vertical  and  the  chords  not  parallel.  The  only  dif- 
ficulty in  constructing  the  moment  influence  diagram  is 
the  determination  of  the  proper  length  of  the  cut  stringer. 
This  once  determined  the  construction  of  the  diagram  follows 
the  general  rule. 

Lower  Segment  of  Diagonal  in  Sub-strut  Truss.  —  Let  VL", 
Fig.  8,  be  the  lower  segment  considered.  This  is  a  portion 
of  the  diagonal  VL"  of  the  main  truss  and  also  forms  a 
chord  member  of  the  auxiliary  truss  LVL". 

Considering  the  member  VL"  as  a  part  of  the  diagonal 
of  the  main  truss,  the  influence  diagram  DD'FBD  is 
constructed  in  the  usual  manner.  The  auxiliary  truss 
LVL"  may  be  considered  as  a  simple  truss  supported  at 
L  and  L"  .  The  center  of  moments  for  VL"  may  be 
taken  anywhere  in  LL"  or  LL"  produced  and  in  order 
that  the  lever  arm  of  VL"  shall  be  the  same  as  that  for 
VL"  the  center  of  moments  will  be  taken  at  the  inter- 
section of  UU"  and  LL"  produced,  the  center  of  moments 
for  VL".  The  influence  diagram,  constructed  according 
to  the  general  rule,  is  D'EBf.  Combining  the  two  diagrams 
just  constructed,  the  final  influence  diagram  for  VL" 
is  DEFBD.  This  diagram  is  the  same  as  if  constructed 


16 


INFLUENCE  DIAGRAMS 


with  the  horizontal  projection  of  VL"  as  the  length  of  the 
cut  stringer.  The  values  of  x'  and  d'  can  now  be  found 
in  the  usual  manner  from  the  truss  diagram. 


The  criterion  for  the  position  of  wheel  loads  producing 
maximum  moment  is  found  in  the  manner  previously  ex- 
plained. Referring  to  equation  (13)  and  Fig.  8. 


dx 


or 


-  =  W2(  —  tan  63  -ftan  02)  —  TF2'(tan  6$  —  tan  d>>] 


tan 


-W3  tan  03=0,  .  (42) 
')  tan  02.    .     (43) 


Substituting  the  values  of  the  tangents 


..     (44) 


SIMPLE  TRUSSES 


17 


and 


which  is  the  required  criterion. 

In  case  the  chords  are  parallel  s  becomes  oo   and  the 
criterion  is 


W    W2+W2' 
l'~         d       ' 


.     (46) 


Upper    Segment   of   Diagonal   in   Sub-strut  Truss. — Refer- 
ring to  Fig.  9,  it  is  clear  that  the  diagonal   UV  has  the 


A  ---ife^S 

" — ^SsSrs 


FIG.  9. 


stress  of  ULf,  the  diagonal  of  the  main  truss,  as  the 
auxiliary  truss  LVL"  simply  serves  the  purpose  of  a 
trussed  stringer  extending  from  L  to  L" .  The  influence 
diagram  is  constructed  according  to  the  general  rule  taking 
the  length  of  the  cut  stringer  as  the  horizontal  projection 


18 


INFLUENCE  DIAGRAMS 


of  LL".     The  graphical  determination  of  df  and  x'  is  evident 
from  the  truss  diagram. 

The    criterion    for    wheel    loads    producing    maximum 
moment  i 


')j.     .     .     (47) 


sd 


If  the  chords  are  parallel  the  criterion  becomes 


(48) 


Vertical  of  Sub-strut  Truss. — As  indicated  in  Fig.  10,  the 
vertical  of  a  sub-strut  truss  is  a  member  of  the  main  truss. 


FIG.  10. 


The  auxiliary  truss  acts  simply  as  a  trussed  stringer.  The 
construction  of  the  influence  diagram  follows  the  general 
rule  using  the  horizontal  projection  of  LL"  as  the  length 


SIMPLE  TRUSSES 


19 


of  the  cut  stringer.     The  graphical   determinations  of  d' 
and  x'  are  clearly  shown  in  Fig.  10. 

The    criterion    for    wheel    loads    producing    maximum 
moment  is 


)         .    .  "(49) 

Ll 


Upper    Segment    of  Diagonal    in    Sub-hanger    Truss.—  The 
member  UV  in  Fig.  11,  is  a  part  of  the  diagonal  UL"  of 


h-h 


//  Tan  0i=l.      Tan  02  =—    .     Tan  63  =— . 

'E  *d  l 


FIG.  11. 


the  main  truss  and  also  a  chord  member  of  the  auxiliary 
truss  UVU".  The  influence  diagram  for  UL"  js  DEF'BD 
and  that  for  the  chord  UV  is  D'FB'.  The  combination 
of  these  two  diagrams  gives  the  influence  diagram  DEFBD. 
This  diagram  is  constructed  according  to  the  general 
rule  when  the  horizontal  projection  of  UV  is  taken  as  the 
length  of  the  cut  stringer.  The  graphical  determination 
of  x'  and  df  requires  no  explanation  as  the  constructions 
are  evident  in  Fig.  11. 


20 


INFLUENCE  DIAGRAMS 


The    criterion    for    wheel    loads    producing    maximum 
moment  is 


W    W2+W2' 
I  ~       d' 


(50) 


Lower  Segment  of  Diagonal  in  Sub-hanger  Truss. — As  in 
the  case  of  the  upper  segment  in  the  sub-strut  truss  this 
member  is  a  part  of  the  diagonal  of  the  main  truss  and  has 


K-h 


.     Tan0i=— . 
»  * 


FIG.  12. 


its  stress  only.  Taking  the  horizontal  projection  of  UL" 
as  the  length  of  the  cut  stringer  the  influence  diagram 
DEFBD,  Fig.  12,  is  constructed  according  to  the  general 
rule.  The  distances  df  and  x'  are  found  in  the  usual  manner. 
The  criterion  for  wheel  loads  producing  maximum 
moment  is 


W 

I 


d' 


(51) 


SIMPLE  TRUSSES 


21 


Vertical  of  Sub-hanger  Truss. — In  Fig.  13,  the  cut  stringer 
for  the  vertical  UL  is  the  horizontal  projection  of  LL', 
since  no  part  of  the  load  on  UL"  is  supported  at  L.  The 
construction  of  the  influence  diagram  follows  the  general 
rule.  In  determining  x'  and  df  from  the  truss  diagram,  h1 
and  the  diagonal  UL'  form  no  part  of  the  truss  proper. 


$-•-:;; 


The    criterion    for    wheel    loads    producing    maximum 
moment  is 


d' 


(52) 


Bottom  Chord  of  Sub-strut  Truss. — The  bottom  chord 
in  Fig.  14  is  a  part  of  the  chord  of  the  main  truss  and  also 
a  part  of  the  chord  of  the  auxiliary  truss  L3FL5.  Con- 
sidering only  the  chord  L3L3  of  the  main  truss  the  influence 
diagram  is  DABD.  Now  considering  the  chord  L^L5 
of  the  auxiliary  truss  L3VL5  and  taking  the  center  of 


22 


INFLUENCE  DIAGRAMS 


moments  at  V,  the  influence  diagram,  constructed  according 
to    the   general   rule,    is    C'EB'F'C'.     The    distance    C'D' 

(£73L3) 
is  laid  off  equal  to  (AK)  ,yj       in  order  that  the  moments 

may  be  equivalent  to  taking  Us  as  a  center  of  moments. 
The  diagram  is  inverted  and  when  combined   with   the 


FIG.  14. 


previous  diagram  the  two  form  the  influence  diagram  for 
FLo,  as  shown  by  the  shaded  area  DAEFBD. 
Referring  to  Fig.  14, 


D'C':EK::AF:KF    or 


but 


therefore 


AF 


KF=d' 


AF=AK- 


SIMPLE  TRUSSES  23 

If  the  line  DA  is  prolonged  until  it  cuts  the  vertical  through 
K  the  ordinate  cut-off  above  K  equals  AK  and  hence  it 
cuts  the  vertical  at  E,  and  EK  in  the  small  diagram  equals 
EK  of  the  large  diagram.  This  shows,  if  the  horizontal 
projection  of  VL5  is  taken  as  the  length  of  the  cut  stringer 
and  an  influence  diagram  constructed  according  to  the 
general  rule,  that  this  is  the  true  influence  diagram  for  VL5. 
The  criterion  for  -wheel  loads  producing  maximum 
moment  is  found  as  follows: 


-=TFi(-tan  0i+tan  03)-pW2(-tan  0i+tan  03) 

+TT2'(tan  0?+tan  03-)  +TF3  tan  03  =0;     (53) 
then 

W  tan  03  =  (Wi  +W2)  tan  0i  -  W2'  tan  0'2.    .     (54) 

Substituting  the  values  of  the  tangents  and  dividing 
through  by  s 


--  KY-^-.    •  •  •  (55) 

If  b-s=d,  then 

W    TF.+TFz-HY. 


I 


(56) 


Top  Chord  of  Sub-strut  Truss.— In  Fig.  15  the  top  chord 
UU'  has  no  double  duty  to  perform  as  it  is  simply  a  chord 
member  of  the  main  truss.  The  influence  diagram  is 
drawn  according  to  the  general  rule  using  the  horizontal 
projection  of  the  chord  UU'  as  the  length  of  the  cut  stringer. 


24 


INFLUENCE  DIAGRAMS 


The    criterion    for    wheel    loads    producing    maximum 
moment  is 


W    Wi+W2 


(57) 


U. £. ^ ^— ___-J 


Criterion,     -y  =  ( TFi  +  Wz)—. 
Tan  0i  =  1.     Tan  03  =4- 

FIG.  15. 

Top  Chord  of  Sub-hanger  Truss. — Referring  to  the  expla- 
nations given  for  Fig.  14,  the  horizontal  projection  of 
UV,  Fig.  16,  is  taken  as  the  length  of  the  cut  stringer 
and  the  influence  diagram  constructed  according  to  the 
general  rule. 

The  criterion  for  wheel  loads  producing  maximum 
moment  is 


W_Wi     W2(s-b) 
I  ~  s  +"     sd      ' 


.     .     .     (58) 


If  s-b=2d,  then 


W     Wi+2W2 


(59) 


SIMPLE  TRUSSES 


25 


Criterion.     If  =  W>  +  W2(s-b)  ^ 

t  S  SCI 


Tan  0i=l.     Tan  02 


FIG.   16. 


.   d 


Tan  63 


Bottom  Chord  of  Sub-hanger  Truss. — This  case  follows  the 
general  rule  for  simple  trusses.  4  The  influence  diagram 
is  shown  in  Fig.  17. 


-d- •> 


Tan  0i  =  1.     Tan  0»  =-.     Tan  0» 


FIG.  17. 


26  INFLUENCE  DIAGBAMS 

The    criterion    for    wheel    loads    producing    maximum 
moment  is 


The  Length  of  the  Cut  Stringer  for  trusses  with  sub- 
struts  or  sub-hangers  of  the  type  considered  above,  can 
be  readily  determined  by  the  following  rules. 

(a)  For  diagonal  members:  whenever  the  section  cuts 
the  segment  which  is  a  part  of  the  main  diagonal  and  also 
a  part  of  the  auxiliary  truss,  the  length  of  the  cut  stringer 
is  the  horizontal  projection  of  the  segment  cut. 

(6)  For  diagonal  members:  whenever  the  section  cuts 
the  segment  which  is  a  part  of  the  diagonal  of  the  main 
truss  and  forms  no  part  of  the  auxiliary  truss,  the  length 
of  the  cut  stringer  is  the  horizontal  projection  of  the  diagonal 
of  the  main  truss. 

(c)  For  chord  members:    whenever  the  section  cuts  the 
chord  which  forms  a  part  of  the  main  truss  chord  and 
also  a  part  of  the  auxiliary  truss,  the  length  of  the  cut 
stringer  for  this  chord  is  the  horizontal  projection  of  the 
diagonal    segment    cut    by    the    section.     Otherwise    the 
length  of  the  cut  stringer  is  the  horizontal  projection  of  the 
chord,  of  the  main  truss,  which  is  cut. 

(d)  For  vertical  members  of  the    main  truss:   whenever 
the  section  cuts  any  portion  of  the  auxiliary  truss  the  length 
of  the  cut  stringer  is  the  horizontal  projection  of  the  diagonal 
of  the  main  truss.     In  case  the  section  does  not  cut  the 
auxiliary  truss  the  length  of  the  cut  stringer  is  the  horizontal 
projection   of  *the   segment   of   the   diagonal   adjacent   to 
the  vertical. 


SIMPLE  TRUSSES 


27 


SPECIAL    CASE 

Diagonal  of  Simple  Truss,  Having  Center  of  Moments  be- 
tween the  Supports. — In  Fig.  18  make  a  section  cutting 
two  chords  and  the  diagonal  UL'.  The  intersection  of 
the  two  chord  members  is  between  the  supports,  a  condi- 
tion which  has  not  obtained  in  any  of  the  previous  examples. 
The  influence  diagram  for  moments  is  constructed  according 


FIG.  18. 

to  the  general  rule  and  is  shown  by  the  shaded  area  in 
Fig.  18.  This  diagram  shows  that  each  and  every  load 
placed  upon  the  span  produces  the  same  kind  of  stress 
in  the  diagonal  UL1 '. 

The   criterion  for   the   position    of   wheel  loads  which 
produce  the  maximum  moment  is  found  as  follows: 


-— 

oX> 


0i+tan  03)+TF2(-tan  02+tan  03) 

+  TF3tan  03=0,  .     .     .     .'    (61) 


28 
or 


INFLUENCE  DIAGKAMS 


W  tan  63  =  Wi  tan  Bl  +W2  tan  02.       .     .     (62) 


Substituting  the  values  of  the  tangents 


W    W1          s-b 


(63) 


where  W  is  the  total  load  on  the  span. 


FIG.  19. 


Vertical  of  Simple  Truss  Having  Center  of  Moments  between 
the  Supports.  —  The  vertical  member  UX,  Fig.  19,  has 
its  center  of  moments  between  the  supports.  The  moment 
influence  diagram  is  constructed  according  to  the  general 
rule  and  is  the  shaded  area  in  Fig.  19.  This  diagram 


SIMPLE  TRUSSES  29 

is  similar  to  that  in  Fig.   18  and  therefore  the  criterion 
locating  wheel  loads  which  produce  the  maximum  moment  is 


CHAPTER   II 

DOUBLE  INTERSECTION   TRUSSES 

Double  Intersection  Trusses  are  made  up  of  two  or 
more  simple  trusses  and  the  influence  diagram  for  any 
member  is  found  by  first  drawing  the  influence  diagram 
for  each  simple  truss  and  then  connecting  these  diagrams 
so  as  to  form  one  diagram.  This  is  shown  in  the  fol- 
lowing examples. 

Top  Chord  of  Whipple  Truss.— The  Whipple  truss  shown 
in  Fig.  20  is  made  up  of  two  simple  trusses  as  indicated 
by  the  full  and  dotted  lines  of  the  truss  diagram. 

The  chord  UzU±  is  a  part  of  the  top  chord  of  both 
trusses.  Considering  it  as  a  top  chord  member  of  the 
truss  shown  by  the  full  lines,  the  center  of  moments  is 
at  Z/4  and  the  length  of  the  cut  stringer  is  L4L6.  The 
influence  diagram  DAFBD  is  constructed  according  to 
the  general  rule.  When  the  member  UzU*  is  considered 
as  a  part  of  the  top  chord  of  the  dotted  truss,  the  center 
of  moments  is  at  Ls,  the  length  of  the  cut  stringer  is  LsZ/7 
and  the  influence  diagram  is  DA'F'B'D' '.  This  diagram 
is  constructed  upon  the  line  DB  of  the  first  diagram  by 
making  D'C"  equal  s'.  Now  a  load  at  L2  does  not  affect 
the  members  of  the  dotted  truss  and  hence  the  ordinate 
ef  in  the  influence  diagram  for  the  truss  shown  by  full 
lines  is  the  correct  moment  for  a  unit  load  at  L2.  A  load 
at  L3  by  similar  reasoning  has  for  its  true  ordinate  e'f 
in  the  influence  diagram  for  the  dotted  truss.  For  a 

30 


DOUBLE  INTERSECTION  TRUSSES 


31 


load  between  L2  and  L3  the  influence  line  is  a  straight 
line  connecting  e  and  e'.  In  a  like  manner  the  other 
panel  points  are  considered  and  the  final  influence  diagram, 
shown  by  the  shaded  area,  obtained.  The  loads  at  LI 
and  LH  are  assumed  to  be  equally  divided  between  the 
two  trusses  and  hence  the  influence  line  passes  midway 
between  the  two  influence  diagrams  for  these  points. 


Span=  I 


Ui       (J-2        Us  5      U4       !U5       !U6       !U 


"1        kj        L2;         L3;        L«     ?L5|       L0|  L7 

C'    i         m  i   V 


FIG.  20. 


The  position  of  wheel  loads  producing  maximum 
moment  is  best  found  by  trial.  A  criterion  can  be  deduced 
but  it  is  too  complicated  for  practical  use. 

Vertical  of  Whipple  Truss.  —  The  vertical  C/4L4  is  a 
member  of  the  truss  shown  by  the  full  lines,  Fig.  21.  The 
influence  diagram  for  vertical  shear  in  the  panel  cut  is 
DEFBD.  For  loads  at  panel  points  L2,  Z/4)  LQ,  etc.,  the 
ordinates  of  this  influence  diagram  are  correct  for  unit 
loads.  The  loads  at  L3,  L5,  Z/?,  etc.,  are  supported  entirely 
by  the  truss  shown  by  dotted  lines  and  hence  do  not  con- 
tribute any  shear  in  the  panel  being  considered,  of  the 


32 


INFLUENCE  DIAGRAMS 


truss  shown  by  the  full  lines.  Therefore,  the  ordinates 
in  the  influence  diagram  directly  below  these  points  are 
zero.  Connecting  these  points  with  the  ends  of  the  ordinates 
which  are  correct,  as  shown  in  Fig.  21,  the  influence  diagram 
shown  by  the  shaded  area  is  obtained.  The  loads  LI 
and  I/ 11  are  assumed  to  be  equally  divided  between  the 
two  trusses. 


FIG.  21. 

Diagonal  of  Whipple  Truss. — The  diagonal  C/2^4,  Fig.- 
22  is  a  member  of  the  truss  shown  by  full  lines  and  its 
influence  diagram  is  constructed  in  a  manner  similar  to 
that  explained  for  the  vertical  U^L^.  The  influence  diagram 
is  shown  by  the  shaded  areas. 

The  position  of  wheel  loads  producing  the  maximum 
vertical  shear  is  found  by  trial. 

Top  Chord  of  Sub-divided  Double  Triangular  Truss.— The 
two  principal  trusses  are  shown  in  Fig.  23  by  full  and 
dotted  lines.  The  top  chord  member  UeUs  forms  a  part 
of  each  truss.  The  influence  diagram  for  U&Us  as  a  member 
of  the  truss  shown  by  full  lines  is  DABD  and  that  for  the 
truss  shown  by  dotted  lines  is  D'A'B'.  Connecting  these 


DOUBLE  INTERSECTION  TRUSSES 


33 


two  diagrams  in  the  manner  explained  for  the  Whipple 
truss  the  influence  diagram  shown  by  the  shaded  area  is 
obtained. 


FIG.  22. 


BB 


FIG.  23. 


The  position  of  wheel  loads  producing  the  maximum 
moment  is  found  by  trial. 

Bottom  Chord  of  Sub-divided  Double  Triangular  Truss. — 
The  bottom  chord  member  L4I/6,  Fig.  24  is  a  part  of  the 


34 


INFLUENCE  DIAGRAMS 


two  simple  trusses  and  also  a  part  of  the  auxiliary  frame 
LiMzLs.  Neglecting  the  auxiliary  truss  L±MzL§,  the 
influence  diagram  is  drawn  in  the  manner  outlined  for  a 
top  chord  member.  This  diagram  is  DED"B"Fri,  etc.,  and 
for  a  load  of  unity  anywhere  between  Z/4  and  L&  the  moment 
is  equal  to  the  ordinate  immediately  below  the  load  between 
the  lines  D"B"  and  DB,  and  the  stress  in  L4Z/6  equals 
this  moment  divided  by  the  depth  of  the  truss.  The 
stress  in  L±L§  as  a  part  of  the  auxiliary  truss  LiMzLs  can 


FIG.  24. 


be  found  by  drawing  the  influence  diagram  for  L^Ls  and 
dividing  the  ordinate  directly  below  the  load  by  the  length 
L5MS.  If  the  scale  of  this  diagram  is  properly  taken 
the  ordinates  may  be  added  directly  to  those  of  the  large 
diagram. 

Following  the  general  rule  for  constructing  influence 
diagrams,  with  D"C",  Fig.  24,  equal  to  s"  =L±L5  multi- 
plied by  LiUi+LsMz,  the  diagram  D"A"B"  is  obtained. 
The  ordinates  are  increased  in  the  same  ratio  as  the  ratio 
of  the  lever  arms  of  L±Ls.  The  shaded  area  is  the  influence 
diagram  for  L^Ls.  This  is  simply  the  proper  combination 


DOUBLE  INTERSECTION  TRUSSES 


35 


of  three  diagrams  for  simple  trusses.  For  the  truss  shown 
C"  falls  in  the  line  CB.  This  method  of  constructing  the 
influence  diagram  requires  no  preliminary  calculations. 

As  in  the  previous  case  the  position  of  wheel  loads 
producing  the  maximum  moment  is  best  found  by  trial. 

Lower  Segment  of  Diagonal  in  Sub-divided  Double  Tri- 
angular Truss. — Selecting  the  member  L^M-s,  Fig.  25,  it  is 
at  once  seen  that  it  forms  a  part  of  the  diagonal  of  the  truss 


E  D"E" 


FIG.  25. 


shown  by  full  lines  and  also  a  part  of  the  auxiliary  truss 
L^MzLe.  The  influence  diagram  for  vertical  shear,  neg- 
lecting the  effect  of  the  auxiliary  truss,  is  DeD"B"mnopBD. 
The  shear  diagram  for  L4M 3  as  a  member  of  the  auxiliary 
truss  is  D"F"B"D".  This  is  constructed  according  to 
the  general  rule  using  the  line  D"B"  as  the  line  DB  in  the 
usual  diagram.  The  influence  diagram  for  L^Ms  now 
becomes  the  shaded  area  in  Fig.  25. 

The  position  of  wheel  loads  producing  the  maximum 
vertical  shear  is  found  by  trial. 

Chords  of  the  "K"  Truss.— The  truss  shown  in  Fig. 
26  is  called  the  "  K  "  truss  since  the  web  members  form 


36 


INFLUENCE  DIAGRAMS 


a  series  of  the  letter  K.  Making  a  section  cutting  U2U3, 
UzMz,  MoLs  and  L2L3,  it  at  once  appears  that  the  upper 
and  lower  chord  members  in  a  panel  have  equal  horizontal 
components  since  the  centers  of  moments  are  in  the  same 
vertical  line. 

The  construction  of  the  influence  diagram  for/  moments 


FIG.  26. 


follows  the   general   rule.     The   criterion  for  wheel   loads 
which  provide  the  maximum  moment  is 


W 


(65) 


where  W  is  the  total  load  on  the  span  and  Wi  the  total 
load  between  the  left  support  of  the  truss  and  the  center 
of  moments  of  the  chords  considered. 

Diagonal  Web  Members  of  the  "  K  "  Truss.— The  usual 
method  of  sections  cannot  be  employed  as  any  section 
made  through  the  truss  cuts  at  least  four  members.  From 
the  arrangement  of  the  diagonals  it  is  clear  that  the  pair 
panel  must  have  horizontal  components  equal  in 


DOUBLE  INTERSECTION  TRUSSES 


37 


magnitude  but  opposite  in  character.  The  magnitude  of 
this  component  is  equal  to  the  difference  between  the 
horizontal  components  of  the  stresses  in  the  chords  in 
the  panel  containing  the  diagonals  being  considered  and  the 
adjacent  panel  opposite  the  point  of  intersection  of  the 
diagonals.  In  Fig.  27  the  horizontal  components  of  the 
diagonals  U^M2  and  M2L%  is  the  difference  between  the 
horizontal  components  of  the  chords  in  this  panel  and  the 
panel  immediately  upon  the  right. 


JLJL 


The  influence  diagram  for  the  horizontal  component 
of  .U$U-±  and  L^L^  can  be  constructed  according  to  the 
general  rule  by  laying  off  CD  equal  to  s+h2.  On  DB  as 
a  base  construct  a  similar  diagram  for  the  chords  U2Uz 
and  L2L3,  the  difference  between  these  diagrams  as  shown 
by  the  shaded  areas  in  Fig.  27  being  the  influence  diagram 
for  the  horizontal  components  of  the  diagonals  U3M2  and 
M2Lz.  The  stress  in  any  diagonal  equals  its  horizontal 
component  multiplied  by  the  secant  of  the  angle  it 
with  the  horzontal. 


38 


INFLUENCE  DIAGRAMS 


The   criterion   for   the   position   of  wheel  loads  which 
produce  maximum  stresses  is  found  as  follows  : 


-Tf72(-tan  03'+tan  02 


'-  tan  02) 

+  TF3tan03'=0,     (66) 


or 


F(tan030=(^2+TF2')tan02.    .     .     .     (67) 
Substituting  the  values  of  the  tangents, 

'):r-^7r-.  (68) 


In  case  the  chords  are  parallel,  then  hi=h<2  and  s—s'=d' 
and  the  criterion  becomes  the  same  as  developed  for  Fig.  4. 

Upper  Segment  of  Vertical  in  the  "  K  "  Truss.— In  Fig.  28 
a  section  cutting  the  upper  segment  of  the  vertical  C73Z>3 


FIG.  28. 

also  cuts  the  members  t/3C/4,  MzLz,  and  I/2l/3.  Taking 
the  center  of  moments  at  the  intersection  of  the  chord 
members  UsU^  and  L^L^  the  moment  influence  diagram 


DOUBLE  INTERSECTION  TRUSSES 


39 


is  constructed  according  to  the  general  rule.  In  this 
case  the  stress  in  U^M^  is  not  equal  to  the  ordinates  of  the 
diagram  divided  by  the  lever  arm  s+b+d,  since  a  portion 
of  each  ordinate  corresponds  to  the  moment  of  the  stress 
in  the  diagonal  M^L^.  The  simplest  way  to  utilize  the 
influence  diagram  is  to  first  determine  the  stress  in  U?Mz 
for  a  unit  load  at  L3.  (This  can  be  very  quickly  done 
graphically.)  Then  below  F  in  the  influence  diagram  lay 


FIG.  29. 

off  to  any  convenient  scale  the  distance  Fg  equal  to  this 
stress.  Draw  a  straight  line  through  B  and  g  and  extend 
it  until  the  point  D  is  located.  Complete  the  diagram 
according  to  the  general  rule.  The  ordinates  in  this 
diagram  correspond  to  the  actual  stresses  in  the  piece 
UsM 3  produced  by  a  unit  load  moving  across  the  span. 

Lower  Segment  of  Vertical  in  the  "  K  "  Truss. — As  in  the 
case  of  the  upper  segment  it  is  impossible  to  make  a  section 
which  does  not  cut  at  least  four  members  of  the  truss. 
Practically  the  same  method  is  employed  as  outlined  for 


40  INFLUENCE  DIAGRAMS 

the  upper  segment.  Graphically  determine  the  stress  in 
MzL3  produced  by  a  unit  load  at  L±  and  lay  off  this  stress 
below  F  as  Fg.  Locate  D  by  drawing  a  straight  line  through 
B  and  g  and  then  complete  the  influence  diagram  according 
to  the  igeneral  rule.  This  gives  the  diagram  DEFBD. 
A  unit  load  at  L3  produces  a  stress  in  M3L3  due  to  its 
position  in  the  principal  truss  and  also  a  local  stress  con- 
sidering M 3L3  as  a  piece  which  simply  transfers  a  portion 
of  the  load  at  L3  to  the  main  truss.  While  the  stress  in 
M3L3  produced  by  a  unit  load  at  Z/3  is  readily  computed 
yet  it  is  usually  more  satisfactory  to  find  this  stress  graph- 
ically by  means  of  an  ordinary  stress  diagram.  After 
this  stress  is  found,  lay  off  mn  equal  to  it  and  connect 
/  and  n  and  F  and  n.  The  shaded  area  shown  is  the 
stress  influence  diagram  for  the  member  M3L3. 


CHAPTER   III 

CONTINUOUS  TRUSSES 

The  Moment  Influence  Diagrams  for  various  forms  of  simple 
trusses  have  been  explained  in  detail  and  found  to  follow 
a  single  general  rule  in  their  construction.  After  the  dis- 
tance of  the  center  of  moments  from  the  left  support  of 
the  span  is  determined  the  diagram  is  constructed  without 
further  calculations  arid  the  ordinates  are  the  moments 
for  unit  loads,  on  the  truss,  directly  above. 

For  trusses  which  are  continuous,  or  partially  con- 
tinuous, the  moment  influence  diagrams  are  founded  upon 
the  influence  diagram  for  simple  trusses.  The  diagram 
for  simple  trusses  will  be  called  the  base  diagram  for  con- 
venience. 

General  Moment  Formula.  —  Considering  any  span  of  a 
continuous  girder,  let 

Ms  =  the  moment  about  a  center  of  moments  distant  s 

from  the  left  end  of  the  span. 

ML  =  the  bending  moment  at  the  left  end  of  the  span. 
MR  =the  bending  moment  at  the  right  end  of  the  span. 
SL  =the  vertical  shear  at  the  left  end  of  the  span. 
W  =  any  concentrated  vertical  load  in  the  span. 
a  =  the  distance  of  W  from  the  left  end  of  the  span. 
Ri  =the  left  reaction  produced  by  W  considering  the  span 

as  a  simple  beam  on  two  supports. 
I  =  the  length  of  the  span. 

41 


42  INFLUENCE  DIAGRAMS 

w,  =  the  moment  about  a  center  of  moments  distant  s 
from   the   left   support   of   the   truss,    considering 
the  truss  as  a  simple  girder  on  two  supports, 
s  =  the  distance  of  the  center  of  moments  from  the  left 
end  of  the  span.     Positive  when  measured  to  the 
right. 


Then 


s>a 

s  =  ML+SLs-W(s-a),     ....     (a) 


but 

MK-M, 


L 
therefore 


/l/f    —  l/f  s>a 

M,=ML+-          Jis+R1s-W(s-a).       .     (c) 


s>a 

Now,  RiS  —  W(s—a)  =  mg, 
hence 

w 


This  expression  shows  that  the  influence  diagram  for 
Ms  may  be  considered  as  being  composed  of  two  separate 
diagrams  combined  algebraically.  One  of  the  diagrams 
is  that  for  the  common  moment  ma  or  the  base  diagram. 
It  will  be  shown  in  the  problems  which  follow  that  the 
influence  diagram  for  Ms  can  be  easily  constructed  upon 
the  base  diagram  and  with  a  small  amount  of  calculation. 


CONTINUOUS  TRUSSES 


43 


Cantilever  *Truss.  —  Truss  bridges  which  are  called 
cantilever  bridges  are  composed  of  anchor  spans,  cantilever 
spans  and  suspended  spans.  The  simplest  form  of  this 
combination  is  shown  in  Fig.  30. 

Top  Chord  of  Anchor  Span. — Assume  any  chord  as  UzU^, 
Fig.  30,  with  its  center  of  moments  at  L2.  For  loads  in 
this  span  the  influence  diagram  is  the  base  diagram  rep- 
resenting the  moments  ms  in  equation  (d),  since  the  moments 
ML  and  MR  are  zero. 


FIG.  30. 


The  criterion  for  wheel  loads  in  this  span  which  produce 
the  maximum  moment  is 


(69) 


For    loads    in  the  cantilever  span  ML=Q,   MR 
and  from  equation  (d), 


M,=  - 


(70) 


44  INFLUENCE  DIAGRAMS 

This  shows  that  the  lines  DB  and  CB  of  the  base  dia- 
gram are  extended  until  they  cut  the  vertical  through  the 
right  end  of  the  cantilever.  The  influence  diagram  is 
shown  by  the  shaded  area  BKL. 

For  loads  in  the  suspended  span, 

ML  =  0,      MR  =  -  W312^12,      ms  =  Q, 
and  hence 

Ms  =  -Wz(h-^Y"f (^1) 

This   is    the    equation    for    the    straight    line   KM  when 

tan  65  =  j~  T~  =  ~T~-     The   shaded    area   in    Fig.    30   is   the 
LI  63       63 

complete  moment  influence  diagram  for  the  chord  UzUz. 
The  criterion  for  wheel  loads  which  produce  the  max- 
imum negative  moment  is 


_ 

12  ""  h  ' 

Top  Chord  of  Cantilever  Span.  —  The  chord  member, 
Ui  U2,  Fig.  31  will  be  considered.  For  loads  in  this  span 
ML—  —  W2d2,  MR=Q  and  hence,  from  equation  (d), 


(72) 


The  base  diagram  is  first  constructed  for  the  term  m,  as 
shown  by  the  heavy  lined  triangle  in  Fig.  31.  Prolong 
the  line  DA  to  K;  then  from  the  triangle  DKB, 

b'c'  :  KB::a2  :  Z2, 


and 


CONTINUOUS  TRUSSES 


2T^r> 
b'c'  =  -KB  = 


45 


But  this  expression  for  b'cf  is  the  same  as  the  coefficient 
of   W2  in  the  expression  for  Ms   and  hence  the  triangle 

DKB  is  the  influence  diagram  for  TF2a2-^i — .     Since  the 

1 2 

ordinates  of  the  base  diagram  are  positive  and  the  ordinates 
of  this  diagram  are  negative,  the  combination  of  the  two 


Tan  0i  =1.     Tan  03  =4.     Tan  0-a  =-^-S. 


FIG.  31. 


diagrams  produces  the  shaded  area  AKB  as  the  influence 
diagram  for  loads  in  the  cantilever  span. 

For  loads  in  the  suspended  span  the  influence  diagram 
is  the  shaded  area  KBM. 

The  criterion  for  wheel  loads  which  produce  the  max- 
imum moment  is 


W2' 


(73) 


46 


INFLUENCE  DIAGRAMS 


Diagonal  of  Anchor  Span. — As  shown  in  Fig.  32,  the 
diagonal  11^2,  has  its  center  of  moments  at  the  inter- 
section of  the  two  cut  chords  UWs  and  L2Z/3.  For  loads 
in  this  span  the  moment  influence  diagram  is  the  base 
diagram  as  shown  by  heavy  lined  shaded  figure  in  Fig.  32. 


Tan  63  =4-     Tan  ft  =T' 4- 
h  &      h 


FIG.  32. 


The  criterion  for  wheel  loads  in  this  span  which  produce 
the  maximum  moment  is 


(74) 


For  loads  in  the  cantilever  span,  ML=Q,  MR=  — 
m,  =0,  and 


=W2a2T.     .    .     .     (75) 


Since  p  is  the  tangent  of  the  angle  #3  the  influence  diagram 
^i 

for  M,  is  the  shaded  area  BKL. 


CONTINUOUS  TEUSSES 


47 


For  loads  in  the  suspended  span  the  influence  diagram 
is  the  shaded  area  KLM. 

The  criterion  for  wheel  loads  which  produce  the  max- 
imum positive  moment  is 


(76) 


FIG.  33. 

Diagonal   of   Cantilever   Span. — Referring  to  Fig.  33,  for 
loads  in  the  cantilever  span  ML  =  —  W2a2,  MB  =  0,  and 


+m. (77) 


The  influence  diagram  for  ms  is  the  base  diagram  shown 
by  the  heavy  lines  in  Fig.  33.     From  the  triangle  DEL, 

n'n  :  EL  ::a2  :  12. 


48  INFLUENCE  DIAGRAMS 

and 

,          ^jyr       CL2,         7  N  h~  S 

nn  —  Y~BL  =  7-  (s  —  12)  =  —  az  —  7  —  . 

12  1^2  12 

Therefore  the  triangle  DEL  is  the  influence  diagram  for 
TPoCte  —  7  —  ,  where  W2  —  1-  Combining  this  'with  the  base 

12 

diagram,  the  shaded  figure  EFBL  is  obtained  as  the  influence 
diagram  for  loads  in  this  span. 

For  loads  in  the  suspended  span  the  influence  diagram 
is  the  shaded  area  ELM. 

The  criterion  for  wheel  loads  producing  the  maximum 
moment  is  found  as  follows: 

Without  stating  the  intermediate  equations 


or 


(78) 


W2'  W3     s-h 

d      s-b-d^s-b-d     Z8    "     '    '     '     (    } 


Its-I2=h, 

W2_W2'+W3 
d        s—b—d 


(80) 


CONTINUOUS    AND    PARTIALLY    CONTINUOUS    TRUSSES    OF 
TWO    EQUAL    SPANS 

Partially  Continuous  Truss. — In  Fig.  34  is  shown  a 
common  form  of  the  revolving  draw-bridge.  The  two 
end  spans  are  equal  and  are  separated  by  an  unbraced 


CONTINUOUS  TRUSSES 


49 


short  span  over  the  turntable.  The  moments  M%  and 
1/3  over  the  turntable  are  assumed  equal  and,  for  con- 
venience, their  values  are  determined  as  if  the  truss  was 
a  beam  of  constant  cross-section.  In  any  case  the  values 
of  M2  and  M3  can  be  found  by  a  more  rigid  method,  if 
desired,  without  increasing  the  labor  of  constructing  the 
influence  diagrams. 


FIG.  34. 


Top  Chord  of  a  Partially  Continuous  Truss. — Referring  to 
Fig.  34,  the  top  chord  U^Ui  has  its  center  of  moments 
at  La.  For  loads  in  this  span, 


L  =  0    and 


but 


where 


50  INFLUENCE  DIAGRAMS 

Therefore,  from  equation  (d), 

_       l  fci-fei3  , 
Dividing  through  by  s, 

T=  ~Wl4+Qn  H 


(82) 


The  influence   diagram  for  -   L  is  represented  by   the 

s 

base  diagram  drawn  with  CD  =  l.  This  is  shown  in  Fig. 
34  by  the  heavy  line  triangle. 

ki  —ki3 
The  value  of    ,     A     is  now  computed  for  the  positions 

4  ~r~\)Tl 

of  Wi=  unity,  corresponding  to  the  panel  points  of  the 
truss.  In  the  truss  shown  ki  has  the  values  0,  1/5,  2/5,  3/5, 

7  73 

4/5,  and  1.     The  corresponding  values  of    Aia     are  laid 

off  as  ordinates  above  the  line  DB  directly  below  the  panel 
points  and  the  ends  of  the  ordinates  connected  by  straight 
lines  forming  the  polygon  Dk'd'm'gB.  The  influence 

M 

diagram  for  — *  is  the  shaded   area   shown  in  the  figure, 
s 

To  obtain  the  value  of  Mt  for  any  particular  load  the  cor- 
responding ordinate  in  the  influence  diagram  is  multi- 
plied by  s. 

For  any  other  center  of  moments  as  1/2  it  is  only 
necessary  to  draw  one  straight  line  as  DA'  to  obtain  the 
influence  diagram. 

In  case  the  center  of  moments  lies  in  a  field  covering 
about  one-fifth  the  span  Zi,  adjacent  to  the  turntable, 
the  line  DA  will  cut  the  polygon  Dd'B  and  thereby  indicate 


CONTINUOUS  TRUSSES 


51 


the  fields  of  loading  in  this  span  which  produce  moments 
of  opposite  character. 

The  position  of  wheel  loads  producing  the  maximum 
moment  is  best  found  by  trial. 

For  loads  in  the  third  span 


4+6n 


and 


.      .     .     .     (83) 


4+6n 
Remembering  that  Ii=l3=l  and  dividing  through  by  s, 


s_  _ 
8   '' 


(84) 


The  influence  diagram  corresponding  to  this  expression 
is  shown  by  the  shaded  area  BKM,  Fig.  34. 

The  position  of  wheel  loads  producing  the  maximum 
moment  is  found  by  trial.  This  position  when  found  will 
remain  constant  for  all  centers  of  moments  in  the  first 
span. 


VALUES  OF 


NCMBER  OF  PANELS  IN  SPAN. 


4 

5 

G 

7 

8         9 

Ll 

.0586 

.048 

.0406    .0350 

.0308 

.0274 

Lo 

.0938 

.084 

.  0740    .  0656 

.0586 

.0527 

L* 

.0820 

.096 

.0937 

.0874 

.0806 

.0740 

L4 

.072 

.0925 

.0962 

.0938 

.0891 

U 

.0637 

.0875 

.0952 

.0960 

Le 

.0568 

.0820 

.0925 

L7 

.0513 

.0767 

L, 

.0466 

52 


INFLUENCE  DIAGRAMS 


The  above  table  contains  the  values  of 


for  unit 


loads  placed  at  the  panel  points  of  trusses  having  from 
four  to  nine   panels.     These   values  multiplied  by 


give  the  ordinates  represented  by 


k-k3 


+  Qn 
If  the  ordinates 


4+6n 
in  any  particular  case  are  laid  off  in  an  inverse  order  the 

resulting  polygon  will  represent  the  expression 


4+6/1 


TIG.  35. 


Diagonal  of  a  Partially  Continuous  Truss. — The  diagonal 
LzUz  in  Fig.  35  has  its  center  of  moments  at  the  inter- 
section of  UoUs  and  L2L3.  Since  the  point  of  intersection 
lies  upon  the  left  of  the  left  end  of  the  span,  s  is  negative. 

For  loads  in  this  span, 


if,— 


and 


m 


(85) 


CONTINUOUS  TRUSSES  53 

tn 
The  influence  diagram  for  -  *  is  indicated  by  the  heavy 

lines  in  Fig.  35.  This  is  the  base  diagram  with  CD  =  unity. 
The  ordinates  represented  by  the  expression  —.  fi  are 
laid  off  above  the  line  DB.  The  influence  diagram  for 

•  is  the  shaded  area  DEGFB.     For  any  other  diagonal 

<s 

the  lines  DE  and  EF}  only  are  changed. 

The  position  of  wheel  loads  producing  the  maximum 
moment  is  found  by  trial.  This  is  easily  done  as  the 
influence  areas  are  nearly  triangular  in  form. 

For  loads  in  the  third  span, 

ML  =  0,     MR  =  -  W9la—~*^*+k**-,     ms  =  0, 

and 

Ms       .r  2/c3-3£32+&33 


4+6n' 


The  influence  diagram  is  indicated  by  the  shaded  area  BKM. 

As  before,  the  position  of  wTheel  loads  is  found  by  trial. 

Continuous  Truss  of  Two  Equal  Spans.  —  If  there  is  no  short 
span  over  the  turntable  n=Q  and  the  only  changes  in  the 
influence  diagrams  given  above  will  be  the  ordinates  to 
the  polygonal  figures.  These  will  be  increased  in  the 
ratio  of  4+6n  to  4. 


CHAPTER   IV 

ARCHES 

Arches  with  open  webs  and  with  solid  webs  will  be 
considered.  The  three-hinged  and  the  two-hinged  arch 
can  be  handled  with  but  little  labor.  The  influence  diagrams 
for  fixed  arches  are  complicated  and  are  best  constructed 
from  computed  ordinates. 

Equation  (d)  is  now  modified  by  the  moment  of  the 
horizontal  thrust.  If  the  lever  arm  of  the  horizontal  thrust, 
H,  is  represented  by  y,  then 


(e) 


Diagonals  in  a  Three-hinged  Arch  with  Open  Web. — The 
frame  diagram  of  a  three-hinged  arch  is  shown  in  Fig.  36 
and  the  influence  diagrams  for  the  diagonals  UzLs  and 
C/3Z/4  are  outlined  in  Figs.  36a  and  366. 

Since  there  can  be  no  bending  moments  at  the  hinges 
over  the  supports,  ML  and  MR  in  equation  (e)  are  zero,  and 


The  influence   diagram  for   — -  is   the   base  diagram  con- 

\y 

54 


ARCHES 


55 


structed  with  CD  =-.    This  is  shown  by  the  area  surrounded 

y 
by  heavy  lines  in  Fig.  36a. 


;-< b >r*~a'->i  w 


FIG.  36. 


The  expression  for  H  is 


-- 

This  indicates  that  H  varies  directly  as  the  distances  of  the 
load  from  the  supports  and  hence  the  influence  diagram 


56  INFLUENCE  DIAGRAMS 

for  H  will  be   a  triangle   between   the   support   and  the 
center  hinge.     Placing  a  unit  load  at  the  center  hinge, 


Lay  off  this  distance  above  DB  directly  below  the 
center  hinge  and  thus  locate  the  point  K.  Draw  KD 
and  KB',  then  the  shaded  area  is  the  influence  diagram 

M 

for   —  *    corresponding   to   the   diagonal    U2LS.     Fig.    366 

y 
is  a  similar  diagram  for  the  diagonal  UsL^. 

For  chord  members  the  same  method  is  employed. 
The  triangle  DKB  is  constant  for  all  influence  diagrams. 

The  position  of  wheel  loads  producing  the  maximum 
moment  is  best  found  by  trial  as  some  of  the  criterions 
become  too  complex  for  easy  application. 

To  illustrate  the  shape  of  one  criterion,  consider  the 
influence  diagram  in  Fig.  366  in  deriving  a  criterion  for 
the  diagonal  £/3L4. 

s  f  05+tan  02)  +  TF2'(+tan  05-tan   02) 


+  W3'  tan  05  +  WY  tan  04  =  0,     .     (88) 
or 

(TPi'+T^'+flYHan  05-(ttY  +  TF2')  tan  02 

+WY  tan  04=0.      .     .     (89) 

Substituting  the  values  of  the  tangents 
(HY  +  HY  +  TfY)|  -  (WS+W,1)*-^  -  W4'(~  -i).      (90) 


AKCHES 


57 


Chord  of  Two-hinged  Arch  with  Open  Web. — As  in  the  case 
of  the  three-hinged  arch  the  moments  ML  and  MR  are 
zero,  and 


y 


(91) 


FIG.  376.      WEB  MEMBERS 
FIG.  37. 


The  influence  diagram  for  --is  the  base  diagram  drawn 

o 

with   CD  equal  -.     This  is  indicated  by  heavy  lines  in 

Fig.  37a.  The  values  of  H  for  a  unit  load  at  each  panel 
point  of  the  loaded  chord  are  computed  and  laid  off  above 
DB  and  the  upper  ends  connected  by  straight  lines.  The 


58 


INFLUENCE  DIAGRAMS 


algebraic  difference  of  these  two  diagrams  is  the  shaded 
area  shown  in  Fig.  37a  and  is  the  influence  diagram  for 

-*  corresponding  to  the  member  LiZ/2. 


FIG.  386.      VERTICAL  SHEAR 

FlG.  38. 


Diagonal  or  a  Two-hinged  Arch  with  Open  Web. — The 
method  followed  in  drawing  the  influence  diagram  for  the 
diagonal  is  the  same  as  explained  for  a  chord  member. 
The  influence  diagram  for  UiL2  is  shown  by  the  shaded 
area  in  Fig.  376. 


ARCHES  59 

The  position  of  wheel  loads  producing  the  maximum 
moment  is  found  by  trial. 

Moment  Influence  Diagram  for  Two-hinged  Arch  with  Solid 
Web. — For  the  solid  arch  the  influence  diagram  is  quite 
simple  as  it  is  made  up  of  the  combination  of  the  base 
diagram  and  the  diagram  for  H,  corresponding  to  unit  loads, 
which  is  a  smooth  curve.  The  moment  influence  diagram 
is  shown  by  the  shaded  area  in  Fig.  38a.  For  any  other 
center  of  moments  than  the  one  indicated  it  is  necessary 
to  change  only  the  lines  CB  and  DA. 

Vertical  Shear  for  Two-hinged  Arch  with  Solid  Web.— The 
influence  diagram  for  vertical  shear  is  the  same  as  explained 
for  simple  trusses  on  two  supports  by  making  the  length 
of  the  cut  stringer  zero  so  that  E  and  F  of  the  base  diagram 
lie  in  the  same  vertical  line.  This  diagram  is  shown  in 
Fig.  386. 


CHAPTER   V 
BEAMS    OF    CONSTANT    CROSS-SECTION 

The  Influence  Diagrams  for  beams  are  but  little  different 
from  those  which  have  been  explained.  In  all  cases  the 
final  diagram  can  be  constructed  upon  the  base  diagram. 

Restrained  Beams. — Referring  to  Fig.  39,  the  moment 
at  any  section  X  is  given  by  equation  (d)  or 


l  —  s 
Ms  =  MR~- 


s 

?r  +w*. 


(d) 


i< 


ywx 


FIG.  39. 
If  the  vertical  shear  at  the  section  X  is  represented  by  SSJ 

s>a        MR— ML  s>? 

s>a 

But  Ri  —  2W=S  =the   vertical  shear  for  a  simple  beam 
resting  upon  two  supports;  therefore 


os  =— 


4-0. 


60 


BEAMS  OF  CONSTANT  CROSS-SECTION 


61 


Simple  Beam  on  Two  Supports. — For  a  simple  beam  rest- 
ing upon  two  supports  ML  and  MR  are  zero,  hence 

Ms  =  ms    and     Ss  =  S. 


FIG.  40. 


The  influence  diagrams  for  ??is  and  $  are  base  diagrams 
constructed  according  to  the  general  rule.  Figs.  40a 
and  406  show  moment  influence  diagrams.  In  one  diagram 
CB  is  drawn  horizontal  according  to  rule,  while  in  the 
other  DB  is  horizontal.  Evidently  the  ordinates  are  the 
same  in  both  figures. 


62  INFLUENCE  DIAGRAMS 

The  influence  diagrams  for  S  are  shown  in  Figs.  40c 
and  40d. 

The  criterion  for  wheel  loads  which  produce  the  maxi- 
mum moment  is 


(93) 


Beam  Fixed  at  One  End  and  Supported  at  the  Other  End. — 

k—k3 
In   this  case  referring  to  Fig.  41,  ML=Q,  MR=  -Wl    „    , 

and 

^=_T7~3+^ (94) 


The  influence  diagram  for  ---  is  the  base  diagram  drawn 

s 

with  CD  -  unity  as  shown  by  the  heavy  lines  in  Fig.  41a- 

k—k3 
The  influence  diagram  for  — -^-~  is  the  figure  DGBD  and 

M 

the   influence    diagram   for   :    -  is  the  shaded  area.     For 

s 

any  other  center  of  moments  the  only  line  changed  is  DA. 
For  vertical  shear 


The  influence  diagram  for  S  is  shown  by  the  heavy  lines 

k—k3 
in  Fig.  416.     The  curve  representing  — ~ —  is  drawn  above 

DB  and  then  the  shaded  area  is  the  influence  diagram 
for  &. 


BEAMS  OF  CONSTANT  CROSS-SECTION  63 

Beam  Fixed  at  Both  Ends.— In  this  case, 


and 


or 


FIG.  41  b.  VERTICAL  SHEAR. 
FIG.  41. 


M,  =  -Wl(k-2k2+k*)l-j----Wl(k2  -k*)j- 


Mt 

s 


.     (96) 


64 


INFLUENCE  DIAGEAMS 


The  base  diagram  indicated  by  the  heavy  lines  in  Fig. 

42a,  is  the  influence  diagram  for  — .     The  ordinates  to 

s 


«,-«  —  .  ,  J 

1     >• 

|w  |w/  s[R 


FIG.  42  6.  VERTICAL  SHEAR 


FIG.  42. 


the  curve  D(75  are  obtained  by  giving  k  corresponding 
values  in  the  expression 


M8 

Then  the  shaded  area  is  the  influence  diagram  for  —  -. 

s 


BEAMS  OF  CONSTANT  CROSS-SECTION 
For  vertical  shear 


65 


(97) 


The  shaded  area  in  Fig.  426  is  the  influence  diagram 
for  Ss. 

VALUES  OF  k*-W  AND  k-2k*+k* 


k 

£2  -A;' 

k 

k*-k* 

0 

0 

1.00 

.05 

.002375 

.95 

.50 

.  125000 

.50 

.10 

.009000 

.90 

.55 

.  136125 

.45 

.15 

.019125 

.85 

.60 

.  144000 

.40 

.20 

.032000 

.80 

.65 

.  147875 

.35 

.25 

.046875 

.75 

.70 

.  147000 

.30 

.30 

.063000 

.70 

.75 

.  140625 

.25 

.35 

.079625 

.65 

.80 

.  128000 

.20 

.40 

.096000 

.60 

.85 

.  108375 

.15 

.45 

.111375 

.55 

.90 

.081000 

.10 

.50 

.  125000 

.50 

.95 

.045125 

.05 

• 

1.00 

0 

0 

k-2k2+k* 

k 

k-2k*+k* 

k 

k-3k*+2k*  =  (k-2k2+k*)-(k*-k*) 

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